Bourbaki

Exercise 7, § 9

Tag 016B, permanent. Cite it as [Bourbaki, Tag 016B].

Lie Groups and Lie Algebras, VIII, 256

Bourbaki marks this one as harder than the rest.

We retain the notations of the proof of Th. 2.

a) Show that

fρ(J(eμ))=∏α∈R+(e(μ∣α)T/2−e−(μ∣α)T/2) f_{\rho}(J(e^\mu)) =\prod_{\alpha\in R_+}(e^{(\mu|\alpha)T/2}-e^{-(\mu|\alpha)T/2})

b) Take (⋅∣⋅)(\cdot | \cdot ) to be the canonical bilinear form ΦR\Phi_R (Chap. VI, §1, no. 12). Show that

18^{18}For more details on this exercise, see: H. FREUDENTHAL, Zur Berechnung der

Charaktere der halbeinfachen Lieschen Gruppen, Proc. Kon. Akad. Wet. Amsterdam, Vol. LVII (1954), pp. 369-376.

fρ(J(eμ))≡dμTN(1+T248(μ∣μ))f_{\rho}(J(e^\mu))\equiv d_\mu T^N(1 +\frac{T^2}{48}(\mu|\mu)) (mod.TN+3R[[T]].T^{N+3}\mathbf{R}[[T]]),

where dμ=∏α∈R+(μ∣α)d_\mu=\prod_{\alpha\in R_+}(\mu|\alpha ).

c) From b) and the equality J(eλ+ρ)=J(e^{\lambda+\rho}) = ch(E).J(eρ).J(e^{\rho}), deduce the formula

∑(μ∣ρ)2\sum(\mu|\rho )^2 dim Eμ=E^\mu= dim E24 (λ∣λ+2ρ)(\lambda |\lambda + 2\rho ).

μ∈P\mu\in P

d) Assume that g\mathfrak{g} is simple. Show that (ρ∣ρ)=(\rho |\rho ) = dim g/24\mathfrak{g}/24. (Apply cc) with λ\lambda the highest root of R, and use Exerc. 3 of §6.)

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